Simplify the following expression and state the condition under which the simplification is valid. You can assume that $z \neq 0$. $q = \dfrac{9(z + 5)}{z} \times \dfrac{2z}{4(z + 5)} $
Solution: When multiplying fractions, we multiply the numerators and the denominators. $q = \dfrac{ 9(z + 5) \times 2z } { z \times 4(z + 5) } $ $ q = \dfrac{18z(z + 5)}{4z(z + 5)} $ We can cancel the $z + 5$ so long as $z + 5 \neq 0$ Therefore $z \neq -5$ $q = \dfrac{18z \cancel{(z + 5})}{4z \cancel{(z + 5)}} = \dfrac{18z}{4z} = \dfrac{9}{2} $